4)Find the exact solutions of each system of equations x^2+y^2=25 and 9y=4x^2
answer= (3,4),(-3,4)
This is correct from what I found after entering the functions into my TI-83 calculator and using the trace=> intersect function on the calculator.
To find the exact solutions of the system of equations x^2+y^2=25 and 9y=4x^2, we can solve them simultaneously.
First, let's solve the second equation for y in terms of x:
9y = 4x^2
y = (4x^2)/9
Now substitute this value of y into the first equation:
x^2 + (4x^2/9)^2 = 25
Simplifying the equation gives us:
x^2 + (16x^4)/81 = 25
Multiplying through by 81 to get rid of the fraction gives us:
81x^2 + 16x^4 = 2025
Rearranging the equation:
16x^4 + 81x^2 - 2025 = 0
This is a quadratic equation in terms of x^2, let's substitute x^2 with a new variable t:
16t^2 + 81t - 2025 = 0
Now, we can solve this quadratic equation for t by factoring, completing the square, or using the quadratic formula. In this particular case, factoring might be difficult, so let's use the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
For our quadratic equation, a = 16, b = 81, and c = -2025. Substituting these values gives us:
t = (-81 ± sqrt(81^2 - 4(16)(-2025))) / (2(16))
Simplifying within the square root:
t = (-81 ± sqrt(6561 + 129600)) / 32
t = (-81 ± sqrt(136161)) / 32
We have two possible values for t:
t₁ = (-81 + sqrt(136161)) / 32
t₂ = (-81 - sqrt(136161)) / 32
Now, we can solve for x by taking the square root of each t value:
x₁ = sqrt(t₁), x₂ = sqrt(t₂), x₃ = -sqrt(t₁), x₄ = -sqrt(t₂)
Finally, substitute the values of x into the equation y = (4x^2)/9 to find the corresponding y values for each x.
By substituting the values of x into the equation y = (4x^2)/9, we find the following solutions for the system of equations:
(x, y) = (3, 4), (-3, 4)